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Vector & Fourier AnalysisPH1202
Part1: Newtonian Space-Time and Vector Analysis
by Dennis Dunn
Version date: Thursday, 16 February 2006 at 12:33Space-TimePhysics–S
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Copyright c©2001 Dennis Dunn.
Contents
Contents 3
Preamble 5
1 Physics 7
2 Newtonian Space-Time 9
2.1 Time 9
2.2 Space 9
3 Spatial Displacements 11
3.1 Distance 12
3.2 Addition 12
3.3 Multiplication by a scalar 12
3.4 Vector Multiplication 12
4 Reference Frames 15
4.1 Some useful vector formulas 16
4.2 Space Points 17
5 Lines, Planes, Spheres et cetera 18
6 Change of reference frame 20
6.1 Vectors in general 23
6.2 Shift of Origin 23
6.3 Galilean Transformations 24
7 Inertial and Non-Inertial Reference Frames 25
7.1 Inertial reference frames 25
7.2 Galilean relativity 25
7.3 Non-inertial reference frames 26
4
8 Differentiation of scalar and vector fields 31
8.1 Fields 31
8.2 Differentiation as an Operator 31
8.3 Vector Differentiation 32
8.4 Differentiation of Scalar Fields 32
8.5 Differentiation of Vector Fields 33
9 Integration of scalar and vector fields 35
9.1 Volume Integrals 35
9.2 Surface Integrals 36
9.3 Line (or path) integrals 37
10 Evaluation of integrals 38
10.1 Volume Integrals 38
10.2 Surface Integrals 40
10.3 Line Integrals 40
11 Vector Integration Theorems 42
11.1 Gauss’s Theorem 42
11.2 Stokes’s Theorem 43
Index 44
Preamble
Aims
The aims of this part of the module are to give you an understanding of the concepts
of space and time.
Intended Learning Outcomes
After the Vector Analysis part of this module a student should be able to:
• discuss the Newtonian concept of space-time;
• discuss how the concept of distance specifies other aspects of geometry;
• describe what is, and what is not, a vector;
• state, and use, the rules for addition and multiplication of vectors;
• integrate and differentiate vector functions of a single variable (eg time);
• determine the relationship between the components of a vector in different reference
frames;
• state the principle of Galilean relativity (the laws of physics have the same form in
all inertial reference frames);
• apply Galilean transformations;
• describe the qualitative effects of non-inertial (accelerating and rotating) reference
frames;
• describe the concepts of scalar and vector fields;
• define and use the vector differential operator−→∇;
• evaluate volume, surface and line integrals;
• state and use the integration theorems of Gauss and Stokes.
Reference Texts
Suitable texts for the first part of the module (vector analysis) are:
Applied vector analysis by Hwei P Hsu (Harcourt Brace Javanovich).
This provides a good coverage of most of subject (there is no treatment of change of
reference frames): its main virtue is that it provides a large number of worked examples.
5
6
Schaum’s outline of Vector Analysis M R Spiegel (McGraw-Hill ISBN 007060228X)
Summary
The syllabus includes the following topics:
Vector Analysis
• Newtonian space-time and invariant distance and time-interval
• Spatial displacements and vectors
• Properties of vectors: addition; scalar and vector multiplication
• Reference frames, basis vectors and vector components
• Change of reference frame and transformation of components
• Galilean transformations; inertial reference frames and Galilean relativity
• Some examples of non-inertial reference frames: Accelerating origin; rotating refer-
ence frames; Coriolis and centrifugal forces (Not examined)
• Concepts of vector and scalar fields
• The vector differential operator ∇• Gradient of scalar field
• Divergence of vector field: significance and definition
• Curl of vector field: significance and definition
• Volume, surface and path integrals of fields: formal definitions and practical methods
of evaluation.
• Integration theorems of Gauss and Stokes.
Assessment
There will be two assessed problem sheets on Vector Analysis and two on Fourier
Analysis issued during the term. These will form 40% of the total module marks.
Such marked assignments provide a means of assessing both each student’s progress and
the progress of the whole class.
The overall understanding developed during the module is assessed through a single
formal University examination and this will form the final 60% of the module assessment.
The examination will be split into two sections: one on vectors and one on Fourier.
Feedback
Feedback is provided by posting solutions to the assessed problems on the Web page.
In addition each student may discuss his/her performance with the lecturer.
Chapter 1
Physics
Experiment or observation is the starting point of any topic in physics. On the
basis of the observations we construct a theory : This is a mathematical model of
(some aspect of ) the physical world. The purpose of a theory is to enable us to make
predictions.
A theory must be internally consistent: That is, none of its postulates should conflict
with any other.
The predictions of a theory should be compared with the results of previous experi-
ments or new experiments should be carried out to test the predictions. If all of these
predictions agree then we should continue to use the theory. If they fail to agree then
we must consider abandoning, or at least, modifying the theory.
From this we can see that it is not possible to prove a theory and we should regard
any theory as suspect.
7
8
This, however, is too simple a view of the way physics works. Unfortunately experi-
mental results have instrumental errors; the predictions of the theories are often based
on approximate, rather than exact, solutions to the mathematical equations. So it is
often not always absolutely clear whether a disagreement between a prediction and an
experimental result should signal the abandonment of the theory.
Chapter 2
Newtonian Space-Time
In this section I set out the Newtonian model or theory that is used for space-time.
We will see later that this theory certainly fails to describe all aspects of physical space-
time: That is, some predictions of the theory fail to agree with results of experiments.
It is never the less a useful model, providing it is used with care.
2.1 Time
We take time to be a one-dimensional mathematical space. That is, a point in this
space is specified by one real variable t .
The unit of time is measured by a device called a clock. The separation between two
times t1 and t2 is called a time-interval and is defined to be |t1 − t2|.
2.2 Space
As a model of space we take a three-dimensional mathematical space. That is, a point in
this space is specified by three real numbers x, y and z. The unit of space is measured
by a device called a ruler .
We choose the three numbers x, y and z to be the distances along three perpendicular
axes. An axis is simply a straight line.
We can define perpendicular and straight line in terms of our ruler. This will be discussed
in class.
The theory is completed by defining the distance between two points (x1, y1, z1) and
(x2, y2, z2) to be
√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 (2.1)
In particular the distance between two neighbouring points (x, y, z) and (x+dx, y+dy,
z+dz ) is
9
2.2 Space 10
ds =√
dx2 + dy2 + dz2 (2.2)
The concept of the ruler and the form of distance define all aspects of the geometry
of the space.
We can, for example, define the concepts of straight line, angle, parallel lines
purely in terms of distance. A straight line between two points can be defined as the
line joining the two points which has the shortest distance . We will discuss angle
and parallel lines in class.
Problems 2.1
(i) Consider how, using only the concept of a ruler, you can define
• a straight line between two points;
• a line perpendicular to another line;
• the angle between two straight lines which intersect;
• parallel lines.
(ii) Suppose A and B are two points in space; AB is the straight line joining them; and
that C is a point on this straight line. Show that the straight line joining C and B
is a segment of AB. (“It is obvious” is not a satisfactory solution!)
(iii) Suppose, as before, that AB is the straight line joining points A and B. Now
suppose that C is variable point and form the straight lines BC and AC. Keeping
the length of BC fixed, allow C to move and choose its position such that AC is as
large as possible.
Show that in this case B lies on the straight line AC.
This is then a way of extending a straight line by an arbitrary length.
Chapter 3
Spatial Displacements
Measurements in physics are usually made relative to a reference frame . For example
the study of the motion of an object through space would involve the measurements of
the projections of the displacement of the object along three perpendicular axes at a
series of times.
The form of the experimental results would depend very much on the choice of refer-
ence frame. Two observers using different reference frames would describe the motion
differently.
However we believe that the physical laws and the processes that result from these
laws do not depend on the choice of reference frame. The use of vectors enables us
to express physical laws and describe physical processes without the need to specify a
reference frame.
The simplest vectors are spatial displacements.
The shortest path from space point P to space point P ′ is called the spatial dis-
placement−−→PP′. If the spatial displacement
−−→QQ′ is the same length as
−−→PP′ and can be
obtained from−−→PP′ by displacing each segment of the path by the same distance (ie they
are parallel) we say
−−→PP′ =
−−→QQ′.
This means that spatial displacements are not localized.
11
3.1 Distance 12
3.1 Distance
The concept of distance is implicit in the above definition, otherwise shortest has no
meaning. It is assumed in the Newtonian model of space-time that this distance is
an absolute quantity. That is, it is a quantity that does not depend on the choice of
measurement axes. We also assume, in this model, that time intervals are absolute
quantities.
3.2 Addition
The result of two successive displacements−→A and
−→B is denoted by
−→A +
−→B . Our model
of space is such that this addition process obeys the rules:
−→A +
−→B =
−→B +
−→A
−→A +
(−→B +
−→C
)=
(−→A +
−→B
)+−→C
You should realise the physics that lies behind these mathematical statements.
The first says that making a displacement−→A followed by a displacement
−→B has exactly
the same effect as making a displacement−→B followed by a displacement
−→A.
The second says that making a displacement−→A followed by the displacement
−→B +
−→C
has exactly the same effect as making a displacement−→A +
−→B followed by a displacement−→
C
3.3 Multiplication by a scalar
The multiplication of a spatial displacement−→A by a (real) number α can be defined as
follows:
If α is positive, then α−→A is in the same direction as
−→A but is α times as long; if α is
negative, then α−→A is in the opposite direction to
−→A and is |α| times as long.
3.4 Vector Multiplication
There are two ways in which two vectors can be multiplied:
Scalar product
−→A •
−→B = |
−→A| |
−→B | cos(ΘAB) =
−→B •
−→A (3.1)
3.4 Vector Multiplication 13
|−→A| and |
−→B | are the lengths of
−→A and
−→B . The term “scalar product” is used because
the result of the multiplication is a scalar quantity .
The scalar product−→A •
−→B can be thought of as the length of the projection of
−→A
along the direction of−→B multiplied by the length of
−→B (or vice versa).
The scalar product has the following properties:
−→A • (
−→B +
−→C) = (
−→A •
−→B) + (
−→A •
−→C) (3.2)
−→A •
−→A = |
−→A|2 (3.3)
−→A •
−→B = 0 ⇒
−→A = 0 or
−→B = 0 or
−→A is perpendicular to
−→B (3.4)
−→A •
−→B = |
−→A| |
−→B | ⇒
−→A is parallel to
−→B (3.5)
−→A •
−→B = −|
−→A| |
−→B | ⇒
−→A is parallel to ( -
−→B) (3.6)
It is important to notice that the scalar product can be expressed entirely in terms of
distance:
−→A •
−→B =
1
2
[|−→A +
−→B |2 − |
−→A|2 − |
−→B |2
](3.7)
Vector product
The result of the vector product of two vectors is itself a vector.
−→A ∧
−→B = |
−→A||−→B | sin ΘAB
−→n = −−→B ∧
−→A (3.8)
3.4 Vector Multiplication 14
ΘAB is the angle between the directions of−→A and
−→B ; and −→n is a vector of unit length
that is perpendicular to both−→A and
−→B . There are however two such directions. The
particular direction of −→n is defined as that of a right-handed screw rotating from−→A
towards−→B .
The vector product obeys the rule:
−→A ∧ (
−→B +
−→C) = (
−→A ∧
−→B) + (
−→A ∧
−→C) (3.9)
(This is not easy to prove!)
3.4.1 Vector product and surface area
The magnitude of the vector product−→A ∧
−→B is the surface area of the parallelogram
defined by−→A and
−→B and its direction is a perpendicular to the surface. The particular
perpendicular is defined by the above right-hand screw rule.
Chapter 4
Reference Frames
A reference frame is defined by an origin O and by three mutually perpendicular axes
OX, OY and OZ. The 3 unit vectors along the three axes will be denoted by −→ex, −→ey
and −→ez .
These vectors have the following properties:
−→ex • −→ex = 1 −→ey • −→ey = 1 ; −→ez • −→ez = 1−→ex • −→ey = 0 −→ey • −→ez = 0 ; −→ez • −→ex = 0
−→ex ∧ −→ex = 0 ; −→ey ∧ −→ey = 0 ; −→ez ∧ −→ez = 0−→ex ∧ −→ey = −→ez ; −→ey ∧ −→ez = −→ex ; −→ez ∧ −→ex = −→ey
(4.1)
The last set of equations defines the axes to be right-handed.
Any spatial displacement can be made from 3 successive displacements along the three
axes.
15
4.1 Some useful vector formulas 16
The result is independent of the order of the displacements. Any displacement vector−→A can therefore be written as :
−→A = Ax
−→ex + Ay−→ey + Az
−→ez (4.2)
−→A = Ax
−→ex + Ay−→ey + Az
−→ez
Ax, Ay and Az are the components of−→A in this reference frame.
Using the properties of the vectors −→ex, −→ey and −→ez , the scalar and vector products can be
expressed in terms of components:
−→A •
−→B = AxBx + AyBy + AzBz (4.3)
−→A ∧
−→B = −→ex(AyBz − AzBy) +−→ey(AzBx − AxBz) +−→ex(AxBy − AyBx) (4.4)
A useful way of remembering the vector product is via a determinant:
−→A ∧
−→B =
∣∣∣∣∣∣∣−→ex
−→ey−→ez
Ax Ay Az
Bx By Bz
∣∣∣∣∣∣∣ (4.5)
4.1 Some useful vector formulas
−→A •
(−→B ∧
−→C
)=−→C •
(−→A ∧
−→B
)=−→B •
(−→C ∧
−→A
)The magnitude of any of these three forms gives (±) the volume of the parallepiped
with sides−→A,−→B and
−→C .
−→A ∧
(−→B ∧
−→C
)=−→B
(−→A •
−→C
)−−→C
(−→A •
−→B
)
4.2 Space Points 17
These formulas can both be proved by expressing each side of the equation in in com-
ponent form.
4.2 Space Points
Spatial displacements can be used to label the points in space. First we choose an origin
O and the any space point P can be uniquely determined (and hence can be labelled)
by the displacement−−→OP.
Therefore there is a one-to-one correspondence between space points and displacement
vectors.
Notice however that the space points themselves are not vectors – they have no direc-
tional properties.
Chapter 5
Lines, Planes, Spheres et cetera
[The contents of this section will not form part of the assessment of the
module] .
I want to show in this section how lines, planes etc can be described using vectors.
Consider first a line which points in the direction of unit vector −→n and which passes
through the point P . If I denote the vector−−→OP by −→p then a general point on the line
can be specified by the distance s from the point P:
−→x (s) = −→p + s−→n (5.1)
If I want an equation for the line, which does not involve the parameter s, then I can
take the vector product of the above equation with respect to −→n :
(−→x −−→p )∧ −→n = 0 (5.2)
This is therefore the equation of a line which goes through −→p in the direction −→n .
Now consider a plane which goes through the point P . I can find two unit vectors −→n1
and −→n2 which lie in the plane and which are perpendicular to each other. A general
point in the plane can be specified by two parameters s1 and s2:
−→x (s1, s2) = −→p + s1−→n 1 + s2
−→n 2 (5.3)
Again I can form an equation for the plane which eliminates the parameters s1 and s2.
First I form the vector product of −→n1 and −→n2:
−→n = −→n 1 ∧ −→n 2 (5.4)
This is a vector which is perpendicular to the plane. I then take the scalar product of
the above equation for −→x (s1, s2) with −→n :
(−→x −−→p )• −→n = 0 (5.5)
18
19
This is therefore the equation of a plane which contains −→p and is perpendicular to the
direction −→n .
Now I consider the equation for a circle. Suppose the circle has radius R, centre −→pand lies in the above plane. That is, the two mutually orthogonal unit vectors −→n1 and−→n2 lie in the plane of the circle. A general point −→x on the circle can be specified by
a parameter ϕ which the angle of the radius, at this point, to the direction −→n1. The
equation can then be written as
−→x (ϕ) = −→p + R cos (ϕ)−→n1 + R sin (ϕ)−→n2 (5.6)
Chapter 6
Change of reference frame
Suppose we know the components of a vector−→A in one reference frame S, how do we
determine its components in some other reference frame S ′ ?
Figure 6.1.
This is important because an experimenter is free to choose whatever reference fame he
likes: There is no guarantee that another experimenter, observing the same event, will
choose the same reference frame.
The origin of these two frames is the same I have separated them for clarity. The vector−→A is the same in the two diagrams but, clearly the components in S – (Ax, Ay, Az) –
and in S ′ – (A′x, A
′y, A
′z) – are different.
I can express the vector−→A in two ways:
−→A = −→exAx +−→eyAy +−→ezAz (6.1)
−→A =
−→e′xA
′x +
−→e′yA
′y +
−→e′zA
′z (6.2)
In order to find A′x in terms of Ax, Ay, Az, I take the scalar product of
−→A with respect
to −→ex′.
This yields two results corresponding to the above two ways of writing−→A:
20
21
Figure 6.2. )
−→A •
−→e′x = Ax(
−→ex •−→e′x) + Ay(
−→ey •−→e′x) + Az(
−→ez •−→e′x)
−→A •
−→e′x = A′
x
(6.3)
Equating these two expressions gives
A′x = Ax(
−→ex •−→e′x) + Ay(
−→ey •−→e′x) + Az(
−→ez •−→e′x) (6.4)
which is the required relationship.
This procedure can be repeated using−→e′y and
−→e′z instead of
−→e′x. The complete set of
results relating the components in S ′ to those in S is:
A′x = Ax(
−→ex •−→e′x) + Ay(
−→ey •−→e′x) + Az(
−→ez •−→e′x)
A′y = Ax(
−→ex •−→e′y) + Ay(
−→ey •−→e′y) + Az(
−→ez •−→e′y)
A′z = Ax(
−→ex •−→e′z) + Ay(
−→ey •−→e′z) + Az(
−→ez •−→e′z)
(6.5)
We can also get the relationships the other way round. That is relating the components
in S to those in S ′.
Ax = A′x(−→e′x •
−→ex) + A′y(−→e′y •
−→ex) + A′z(−→e′z •
−→ex)
Ay = A′x(−→e′x •
−→ey) + A′y(−→e′y •
−→ey) + A′z(−→e′z •
−→ey)
Az = A′x(−→e′x •
−→ez) + A′y(−→e′y •
−→ez) + A′z(−→e′z •
−→ez)
(6.6)
22
I leave this as an exercise for you to prove.
What is the significance of the terms like (−→e′y •
−→ez)?
From the definition of scalar product, (−→e′y •
−→ez) is simply the cosine of the angle between
the direction of−→e′y and −→ez (because these vectors have length 1). This is called a
direction cosine .
Hence the above relationships only depend on the relative orientations of the two sets
of vectors−→e′x,−→e′y ,−→e′z and −→ex,
−→ey ,−→ez .
Figure 6.3.
A simple example is the case in which S and S ′ have one axis in common, say−→e′z = −→ez
The direction cosines for this example are:
(−→ex •−→e′x) = cos(θ); (−→ey •
−→e′x) = cos(π
2− θ); (−→ez •
−→e′x) = 0
(−→ex •−→e′y) = cos(π
2+ θ); (−→ey •
−→e′y) = cos(θ); (−→ez •
−→e′y) = 0
(−→ex •−→e′z) = 0; (−→ey •
−→e′z) = 0; (−→ez •
−→e′z) = 1
(6.7)
These can be simplified using cos(π2− θ) = sin(θ) and cos(π
2+ θ) = − sin(θ) .
Hence the complete sets of relations between these two frames are
A′x = Ax cos(θ) + Ay sin(θ)
A′y = −Ax sin(θ) + Ay cos(θ)
A′z = Az
(6.8)
Ax = A′x cos(θ)− A′
y sin(θ)
Ay = A′x sin(θ) + A′
y cos(θ)
Az = A′z
(6.9)
The expressions for these relations between the components are often presented in matrix
form:
6.1 Vectors in general 23
A′x
A′y
A′z
=
−→e′x •
−→ex
−→e′x •
−→ey
−→e′x •
−→ez−→e′y •
−→ex
−→e′y •
−→ey
−→e′y •
−→ez−→e′z •
−→ex
−→e′z •
−→ey
−→e′z •
−→ez
Ax
Ay
Az
(6.10)
Ax
Ay
Az
=
−→ex •
−→e′x
−→ex •−→e′y
−→ex •−→e′z
−→ey •−→e′x
−→ey •−→e′y
−→ey •−→e′z
−→ez •−→e′x
−→ez •−→e′y
−→ez •−→e′z
A′
x
A′y
A′z
(6.11)
The 3×3 matrices of direction cosines are called rotation matrices.
6.1 Vectors in general
So far only spatial displacement vectors have been considered. However the concept of
a vector can be generalized.
The general definition is as follows:
An entity−→G is a vector if
• in any reference frame S it is specified by three component Gx, Gy, Gz which have
the same physical dimensions;
• the corresponding components G′x, G′
y, G′z in another reference frame S ′ are related
to Gx, Gy, Gz by
G′x = Gx(
−→ex •−→e′x) + Gy(
−→ey •−→e′x) + Gz(
−→ez •−→e′x)
G′y = Gx(
−→ex •−→e′y) + Gy(
−→ey •−→e′y) + Gz(
−→ez •−→e′y)
G′z = Gx(
−→ex •−→e′z) + Gy(
−→ey •−→e′z) + Gz(
−→ez •−→e′z)
(6.12)
If−→G satisfies both of these conditions then it can be written as
−→G = Gx
−→ex + Gy−→ey + Gz
−→ez−→G = G′
x
−→e′x + G′
y
−→e′y + G′
z
−→e′z
(6.13)
Velocity, momentum, force, electric and magnetic fields are all examples of vectors.
6.2 Shift of Origin
If two reference frames differ only by a shift of origin to O′ instead of O then this does
not affect spatial displacement vectors or vectors in general. However it does affect the
way in which the space-points are represented by vectors (see section 4.2).
The vectors which represent space points relative to the two origins are related by
6.3 Galilean Transformations 24
−→x′ = −→x −
−−→OO′ (6.14)
where−−→OO′ is the displacement vector from O to O′.
6.3 Galilean Transformations
Suppose that the origin O′ is moving relative to origin O with velocity−→V . Then the
relation between the two vectors −→x and−→x′ describing a point P is
−→x′ = −→x −
−→Vt (6.15)
assuming that the two origins coincide at time t = 0.
This relation holds even if the point P is itself moving.
The velocities of the moving point P relative to the two origins is given by
−→v′ = −→v −
−→V (6.16)
These two vector equations which relate the positions and velocities of a point in two
reference frames moving uniformly with respect to each other are known as the Galilean
transformations
Chapter 7
Inertial and Non-Inertial Reference Frames
A reference frame is defined by three mutually perpendicular unit vectors −→ex, −→ey ,−→ez and
an origin O. It has been implicitly assumed up to now that −→ex, −→ey ,−→ez are independent
of time (although we have allowed O to move).
However these quantities can be chosen to vary with time. One reason for wanting to
do this is that our natural reference frame – the earth – is moving and rotating.
7.1 Inertial reference frames
An inertial reference frame is one in which the origin is moving with constant speed
and in which −→ex, −→ey ,−→ez do not vary with time. That is
−→AO = 0
d−→ex(t)
dt= 0;
d−→ey(t)
dt= 0;
d−→ez(t)
dt= 0
(7.1)
where−→AO is the acceleration of the origin.
7.2 Galilean relativity
Law of relativity: The laws of physics must be the same in any inertial
reference frame.
The means of changing from one inertial reference to another is provided by the above
Galilean transformations.
This law of relativity says that if we apply such a transformation to an equation de-
scribing a physical law then the resulting transformed equation should be look the same
(in terms of the transformed variables) as the original equation.
In other words physics provides no means of distinguishing between two inertial reference
frames.
25
7.3 Non-inertial reference frames 26
7.3 Non-inertial reference frames
[ Only the qualitative aspects of the material on non-inertial reference
frames will be examined. ]
A non-inertial reference frame is one in which one, or more, of the equations (7.1) is not
true.
7.3.1 Moving origin
Figure 7.1.
Consider the small time interval t to t + δt. Suppose that in this time interval the
origin moves from O(t) to O(t+δt), the displacement vector being−→δO(t). Suppose also
that the point P moves from P (t) to P (t + δt), the displacement vector being−→δx(t).
The vectors which label the point relative to the origin are −→x (t) and−→δx(t + δt).
These four vectors are related by:
δ−→x (t) = −→x (t + δ t)−−→x (t) +−→δO(t). (7.2)
The velocity of a point P is defined to be
−→v (t) =δ−→x ( t)
δ t(7.3)
in the limit δt→ 0.
In general, the derivative of any vector function of time, say−→A(t) is defined to be
d−→A(t)
dt=
−→A(t + δ t)−
−→A(t)
δ t(7.4)
7.3 Non-inertial reference frames 27
in the limit δt→ 0.
If we divide the above equation for−→δx(t) by δt and take the limit we get:
−→v (t) =d−→x (t)
dt+−→VO(t) (7.5)
−→VO(t) is the velocity of the origin. In this equation −→v (t) is the actual velocity of
the moving point (a particle say);d−→x (t)
dtis the apparent velocity. This apparent
velocity is what an experimenter would record as the velocity if he did not realise the
point he had chosen as the origin was itself moving.
We can take the equation one stage further by differentiating again:
−→A(t) =
d2−→x (t)
dt2+−→AO(t) (7.6)
Similar interpretations apply to the accelerations:−→A(t) is the actual acceleration of
the moving point;−→AO(t) is the acceleration of the moving origin; and
d2−→x (t)
dt2is the
apparent acceleration that would be measured by an experimenter who neglected the
motion of the origin.
If we insert this expression for the acceleration into Newton’s equation for the motion
of a single particle we get
md2−→x (t)
dt2=−→F (−→x (t))−m
−→AO(t) (7.7)
In terms of the measured (apparent) acceleration there appears to be an additional force
[−m−→AO(t)]. This is called an inertial force and results solely from the motion of the
origin.
7.3.2 Rotating reference frames
The general motion of a reference frame is a translation of the origin, which we have
considered above, and a rotation of the axes about the origin.
Suppose, for the moment, that the reference frame is rotating at an angular rate ω
(radians/sec) about an axis in the direction of unit vector−→n .
As time varies the end-points of the vectors −→ex, −→ey , and −→ez move round the axis−→n in
circular orbits as shown in the diagram. Suppose we concentrate on one of these vectors,
say −→ez , and consider the changes that occur in going from time t to time t + δt.
In this interval the end-point of the vector moves from −→ez(t) to −→ez(t+δt). This end-point
moves on a circle of radius sin(θ) where θ is the angle between −→ez(t) and−→n .
7.3 Non-inertial reference frames 28
Figure 7.2.
The rotation angle in time δt is ωδt. Hence the magnitude of the change is
|δ−→ez(t)| = sin(θ)ω δ t (7.8)
The direction of−→δez(t) is perpendicular to both −→ez(t) and
−→n . Hence
−→δ ez(t) =
(−→n ∧ −→ez(t)
)ω δ t (7.9)
The angular velocity −→ω is defined to be the vector
−→ω = ω−→n (7.10)
its direction is the axis of rotation−→n and its magnitude is the rate of angular rotation
ω.
Taking the limit as δt→ 0 gives
d−→ez(t)
dt= −→ω ∧ −→ez(t) (7.11)
This equation is still valid even if the vector −→ω varies with time. That is, even if the
direction or rate of rotation varies with time.
The complete set of rate equations for the unit vectors in a reference frame rotating
with angular velocity −→ω (t) is
d−→ex(t)
dt= −→ω (t) ∧ −→ex(t) ;
d−→ey(t)
dt= −→ω (t) ∧ −→ey(t) ;
d−→ez(t)
dt= −→ω (t) ∧ −→ez(t) (7.12)
7.3 Non-inertial reference frames 29
Now consider any time-dependent vector−→A(t). It can be written in terms of components
in a rotating reference frame as:
−→A(t) = Ax(t)
−→ex(t) + Ay(t)−→ey(t) + Az(t)
−→ez(t) (7.13)
Two sorts of terms arise when this is differentiated with respect to time:
d−→A(t)
dt=
[dAx(t)
dt−→ex(t) +
dAy(t)
dt−→ey(t) +
dAz(t)
dt−→ez(t)
]+
[Ax(t)
d−→ex(t)
dt+ Ay(t)
d−→ey(t)
dt+ Az(t)
d−→ez(t)
dt
] (7.14)
The first bracket can be used to define a vector:
−→A(t) =
[dAx(t)
dt−→ex(t) +
dAy(t)
dt−→ey(t) +
dAz(t)
dt−→ez(t)
](7.15)
This is the apparent rate of change of the vector: This is what an experimenter
moving with the reference frame would measure.
The complete derivative, using the expressions for the rate of change of the basis vectors,
is
d−→A(t)
dt=−→A(t) +−→ω ∧
−→A(t) (7.16)
This is valid for any vector.
We can go one stage further and evaluate the second derivative:
d2−→A(t)
dt2=−→A(t) + 2−→ω (t) ∧
−→A(t) +−→ω (t) ∧
(−→ω (t) ∧−→A(t)
)+ −→ω (t) ∧
−→A(t) (7.17)
−→A(t) is the apparent second derivative and is what would be measured by an experi-
menter moving with the rotating reference frame. The expression for−→A(t) is
−→A(t) =
[d2Ax(t)
dt2−→ex(t) +
d2Ay(t)
dt2−→ey(t) +
d2Az(t)
dt2−→ez(t)
](7.18)
7.3 Non-inertial reference frames 30
7.3.3 Newton’s Equations in Non-Inertial Reference Frames
If I use the results for rotating reference frames and for accelerating origins then New-
ton’s equations can be expressed as
m−→x (t) =−→F (t)−m
−→A0 (t)
+2 m(−→x (t) ∧ −→ω (t)
)−m−→ω (t) ∧
(−→ω (t) ∧ −→x (t))
− m −→ω (t) ∧ −→x (t)
(7.19)
All the terms on the right-hand side except−→F (t) are called inertial forces: They all a
consequence of using a non-inertial reference frame.
2m(−→x (t) ∧ −→ω (t)
)This is called the Coriollis Force
−m−→ω (t) ∧(−→ω (t) ∧ −→x (t)
)This is the centrifugal force
−m −→ω (t) ∧ −→x (t) This force, which only occurs if the angular velocity is changing, has
no special name.
In the case of a reference frame attached to the Earth |ω| = 7.29 × 10−5 rads s−1 and
ω (t) ≈ 0 There is, however, a slow periodic precession of the axis of rotation with a
period of about 26000 years.
At the surface of the Earth the measured value of g, the gravitational acceleration,
includes the effects of the centrifugal term and the acceleration of the origin–this means
that the vertical does not quite point to the centre of the Earth.
Hence near the surface of the Earth Newton’s equation in the Earth’s reference frame is
m −→x (t) =−→F (t) + m−→g + 2 m
(−→x (t) ∧ −→ω (t))
(7.20)
where −→g includes the gravitational acceleration and the effects of the moving origin
and the centrifugal acceleration and−→F (t) represents all the other forces.
7.3.4 Problem
Explain why −→g includes other terms in addition to the pure gravitational attraction.
Consider how you would weigh something.
Chapter 8
Differentiation of scalar and vector fields
8.1 Fields
A field is a physical quantity defined at every point throughout space.
The simplest type of field is the scalar field : At each point of space a number (or
scalar) specifies the field. An example is the density of the air ρ(−→x ). This is varies from
point to point in space and at each point is defined by a single number (with dimensions
kg m−3 ). The charge density in a charged gas (ie a plasma) is another example.
Note that I use a vector −→x to denote a general space point. This, of course, is relative
to some origin and has components (x, y, z) in a particular reference frame.
Another type of field is a vector field . In this case at each point of space, a vector
specifies the field. An example is the wind velocity −→v (x) in the atmosphere: At each
point in space this is defined by a magnitude (with dimensions m s−1) and a direction.
Electric and magnetic fields; gravitational force fields; current density are all examples
of vector fields.
8.2 Differentiation as an Operator
Before considering fields I want to revert to ’normal differentiation’ and I want to present
this as an operator.
If I consider a function on one variable f (t) then an operator O is simply maps the
function onto another function:
O [f (t)] = g (t) (8.1)
Some simple examples are:
O1 [f (t)] = f (t− t0)
O2 [f (t)] = f (t)2
O3 [f (t)] = 2 f (t)
(8.2)
31
8.3 Vector Differentiation 32
I will now define the operatord
dt:
d
dt[f (t)] = lim
δ t→0
f (t + δ t)− f (t)
δ t(8.3)
The usefulness of this operation really lies in the approximate relation:
f (t + δ t) = f (t) + δ td
dt[f (t)] (8.4)
That is, it tells us how much a function changes when we make a small change in the
argument.
The following properties follow directly from this definition:
d
dt[c f (t)] = c
d
dt[f (t)]
d
dt[f (t) + g (t)] =
d
dt[f (t)] +
d
dt[g (t)]
d
dt[f (t) g (t)] = f (t)
d
dt[g (t)] + g (t)
d
dt[f (t)]
d
dt[f (g (t))] =
d
dg[f (g)]
d
dt[g (t)]
(8.5)
You should try to prove these statements.
8.3 Vector Differentiation
I now revert to the study of fields and define a vector operator−→∇. I will call this
operator del however some texts refer to it as nabla.
−→∇ = −→ex
d
dx+−→ey
d
dy+−→ey
d
dz(8.6)
This is clearly a vector operator as evidenced by the three vectors on the right-hand
side.
8.4 Differentiation of Scalar Fields
I first consider the operation of−→∇ on scalar fields.
8.5 Differentiation of Vector Fields 33
−→∇Φ
(−→x )= −→ex
d
dxΦ
(−→x )+−→ey
d
dyΦ
(−→x )+−→ey
d
dzΦ
(−→x )(8.7)
Each of the differential operators on the right-hand side is a ’normal’ differential oper-
ator.
Note: In performing this operation there is no problem is regarding the field Φ to be a
function of the three variables x, y and z providing you remember that if you change
reference frame then these component values will change (even though the vector −→xdoes not).
This vector operation is also called gradient.
The following properties are a direct consequence of the definition:
−→∇
[c Φ
(−→x )]= c
−→∇Φ
(−→x )−→∇
[Φ
(−→x )+ Ψ
(−→x )]=−→∇Φ
(−→x )+−→∇Ψ
(−→x )−→∇
[Φ
(−→x )Ψ
(−→x )]= Ψ
(−→x ) −→∇Φ
(−→x )+ Φ
(−→x ) −→∇Ψ
(−→x )(8.8)
You should attempt to verify these relations.
8.5 Differentiation of Vector Fields
There are (at least) two ways in which the vector operator−→∇ can be applied to a vector
field.
The simplest involves taking the scalar product of the operator and the vector field:
−→∇ •
−→A
(−→x )=
dAx
(−→x )d x
+dAy
(−→x )d y
+dAz
(−→x )d z
(8.9)
This is pronounced ’del dot A’. It is also called the divergence of−→A
(−→x ).
It is important to note the result of this operation is a scalar field: the result has no
directional properties.
A more complicated form of derivative involves taking the vector product of the operator
and the vector field:
8.5 Differentiation of Vector Fields 34
−→∇ ∧
−→A
(−→x )= −→ex
(dAz
d y− dAy
d z
)
+−→ey
(dAx
d z− dAz
d x
)
+−→ez
(dAy
d x− dAx
d y
)(8.10)
This is pronounced ’del cross A’. It is also called the curl of−→A
(−→x ).
It is important to note the result of this operation is a vector field: it definitely has
directional properties.
8.5.1 Vector Derivative Identities
Here are some useful relationships:
−→∇ •
(−→∇ ∧
−→A
)= 0
−→∇ ∧
−→∇Φ = 0
−→∇ (Φ Ψ) = Φ
−→∇ (Ψ) + Ψ
−→∇ (Φ)
−→∇ •
(Φ−→A
)= Φ
−→∇ •
−→A +
−→A •
−→∇Φ
−→∇ •
(−→A ∧
−→B
)=−→B •
−→∇ ∧
−→A −
−→A •
−→∇ ∧
−→B
−→∇ ∧
(Φ−→A
)= Φ
−→∇ ∧
−→A +
−→∇Φ ∧
−→A
(8.11)
These can all be proved by writing each side of the identity in terms of components.
Chapter 9
Integration of scalar and vector fields
I now consider the integration of scalar and vector fields.
There are three main types of integral:
9.1 Volume Integrals
The volume integrals of vector and scalar fields are denoted by∫V
dV−→A(−→x )
∫V
dV ρ(−→x ) (9.1)
The subscript V is the name given to the volume over which the integration takes place.
The definition of such a volume integral is
lim
N →∞ ; δVn → 0
N∑n=1
δVn
−→A(−→xn) (9.2)
In this expression the whole volume V has been divided into N small volumes δV1 . . . δVN
and −→x n is some point inside the small volume δVn . If the limit exists (and there are
cases where it does not) then we call the result the volume integral and denote it by
(9.1).
As an example, suppose we know the mass density of a material ρ(−→x ) then the total
mass contained in a volume V is
MV =
∫V
dV ρ(−→x ) (9.3)
Of course, in the special case where the density is constant, this just reduces to ρV .
35
9.2 Surface Integrals 36
9.2 Surface Integrals
There are several kinds of surface integral, the most important being denoted by∫S
−→dSΦ(−→x )
∫S
−→dS •
−→A(−→x )
∫S
−→dS ∧
−→A(−→x ) (9.4)
These are defined, for example, by
limN→∞ |
−−→δSn|→0
N∑n=1
−→δSn •
−→A(−→xn) (9.5)
In this case the surface S is divided into N small surface elements.−−→δSn is a vector whose
magnitude denotes the surface area of the nth surface element; −→xn is some point on this
surface. The direction of−−→δSn can be taken to be the perpendicular to the nth surface
element at the point −→xn.
It is important for the definition that all these directions are on the same face of the
surface.
In the case of a closed surface it is conventional to use the outward perpendiculars and
such an integral is denoted by, for example,
∮S
−→dS •
−→A(−→x ) (9.6)
As an example consider electric current. The electric current density in a medium is
−→J
(−→x )= ρ
(−→x ) −→v (−→x )(9.7)
where −→v(−→x )
is the velocity field and ρ(−→x )
is the charge density. The current I crossing
some surface S is
I =
∫S
−→dS •
−→J
(−→x )(9.8)
If the medium is a conducting copper wire then S would the cross-sectional surface of
the wire.
Precisely the same formulas holds if ρ(−→x )
were a mass density for example in a fluid.
The total current flow in a river would be evaluated by an integral like (9.8) where in
this case the surface is the cross-section of the river.
9.3 Line (or path) integrals 37
9.3 Line (or path) integrals
I denote the three most important line integrals by
∫L
−→dl Φ(−→x )
∫L
−→dl •
−→A(−→x )
∫L
−→dl ∧
−→A(−→x ) (9.9)
Again, like a surface, a line (or path) L has directional properties. The formal definitions,
for example, are
limN→∞ |
−→δln|→0
N∑n=1
−→δln •
−→A(−→xn) (9.10)
The (directed) line L is divided into N line segments.−→δln is a vector whose magnitude
denotes the length of the nth segment; −→xn is some point on this segment and the direction
of−→δln is the tangent to the segment at the point −→xn.
If a particle is acted upon by a force field−→F (−→x ) then the work done in moving the
particle along a path L is
WL =
∫L
−→dl •
−→F (−→x ) (9.11)
There a special notation to indicate a closed path:
∮L
−→dl •
−→F (−→x ) (9.12)
Chapter 10
Evaluation of integrals
10.1 Volume Integrals
A volume is three-dimensional: This means that points inside the volume can be de-
scribed by three parameters u, v, and w. The vector labelling the space points can
therefore be regarded as a function of u, v, and w:
−→x = −→x (u, v, w) (10.1)
The question we have to answer is: How can we specify the volume element dV in terms
u, v, and w?
Consider first the simple case where the three parameters are the co-ordinates along
three perpendicular axes x, y and z. In this case we could construct a volume element
by starting at point (x, y, z) and making small changes in each co-ordinate:
This construction produces a rectangular block with sides parallel to the three axes and
the volume is
dV = dxdydz (10.2)
A volume integral could then be evaluated, for example, as
38
10.1 Volume Integrals 39
∫V
dV ρ(−→x ) =
∫dx
∫dy
∫dz ρ(x, y, z) (10.3)
The limits for each of the three integrations need to be carefully determined and de-
pend on the shape of the volume and on the order in which the three integrations are
performed.
If the total volume V is also a rectangular block with sides parallel to the x, y and z
axes then these limits of integration are very simple.
Now we return to the general case in which the vector −→x is specified by 3 parameters
u, v and w. We can construct the volume element in the same way: by make small
changes in each parameter in turn.
The volume of this infinitesimal region is
dV =∣∣∣−−→dxu •
(−−→dxv ∧
−−→dxw
)∣∣∣ (10.4)
where, for example,−−→dxu is −→x (u + du, v, w)−−→x (u, v, w). From this definition, we have
−−→dxu =
∂−→x∂u
du (10.5)
dV =
∣∣∣∣∂−→x∂u•
(∂−→x∂v
∧ ∂−→x∂w
)∣∣∣∣ dudvdw (10.6)
A complete expression for a volume integral is, for example,∫V
dV ρ(−→x ) =
∫du
∫dv
∫dwρ(−→x (u, v, w))|∂
−→x∂u
•(
∂−→x∂v
∧ ∂−→x∂w
)| (10.7)
These are three ‘normal’ integrals. The limits of integration have to be determined from
the shape of the volume.
10.2 Surface Integrals 40
10.2 Surface Integrals
A surface is inherently two-dimensional: This means that it needs to parameters to
specify each point on the surface. I shall call these u and v.
This means that that if a general point on the surface is −→x then this is a function of u
and v:
−→x → −→x (u, v) (10.8)
I can generate the surface element−→dS by making successive changes in u and v.
Figure 10.1. )
−→dS =
(−−→dxu ∧
−−→dxv
)=
(∂−→x∂u
∧ ∂−→x∂v
)dudv
(10.9)
A typical surface integral can the be written as
∫S
−→dS •
−→A(−→x ) =
∫du
∫dv−→A
(−−−−→x(u, v)
)•
(∂−→x∂u
∧ ∂−→x∂v
)(10.10)
Again these are now ‘normal’ integrals. The limits of each integration are determined
by the shape of the surface.
10.3 Line Integrals
A (directed) line or path is one-dimensional. It can be specified by one parameter – u.
10.3 Line Integrals 41
A general point on the line can be represented (relative to some origin) by a vector
function of u: −→x (u).
The line element can be generated by making a small change in u:
Figure 10.2. )
The line element−→dl can then simply be written as
−→dl =
d−→xdu
du (10.11)
and so an integral can be evaluated as
∫L
−→dl •
−→A(−→x ) =
∫du
(d−→xdu
)•−→A
(−→x (u))
(10.12)
The limits of the ‘normal’ integration are the values of u at the start and end of the
line.
Chapter 11
Vector Integration Theorems
There are two important integration theorems:
11.1 Gauss’s Theorem
This theorem relates a volume integration to an integration over the surface of the
volume.
∮S
−→A(−→x ) •
−→dS =
∫V
−→∇ •
−→A(−→x ) dV (11.1)
The direction of−→dS is taken to be pointing out of the surface.
Normally this theorem used to replace the surface integration by the volume integration.
There is no guarantee that this will be simpler but there are cases where it is.
11.1.1 Example
Consider the case in which we need to evaluate
I =∮S
−→A(−→x ) •
−→dS
−→A(−→x ) = x3−→ex + y3−→ey + z3−→ez
(11.2)
where S is a sphere with radius R and with its centre on the origin.
The derivative−→∇ •
−→A(−→x ) is
−→∇ •
−→A(−→x ) = 3 x2 + 3 y2 + 3 z2 (11.3)
and so the right-hand side of Gauss’s theorem is
42
11.2 Stokes’s Theorem 43
I = 3
∫V
(x2 + y2 + z2
)dV (11.4)
The volume is a sphere and we can exploit the spherical symmetry in two ways: First we
can write the integrand as r2 where r is the radius. Second the volume can be integrated
as shells of thickness dr and surface area 4π r2. This gives
I = 12 π
R∫0
r4 dr =12 π R5
5(11.5)
11.2 Stokes’s Theorem
This theorem relates a surface integral to line integral over the line which is the boundary
of the surface.
∮L
−→A(−→x ) •
−→dl =
∫S
(−→∇ ∧
−→A(−→x )
)•−→dS (11.6)
L is a closed path and S is any surface (and there are an infinity of such surfaces) which
has L as its boundary. The direction in which L is taken related to the direction of the
surface is via a right-hand screw rule.
This is normally used to replace the line (or path) integral by (possibly simpler) surface
integral.
11.2.1 Example
11.2 Stokes’s Theorem 44
